STRUCTURE OF TIN ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Tin (Sn) is the element with the greatest number of stable isotopes (ten with atomic masses 112, 114, 115, 116, 117, 118, 119, 120, 122, and 124) (three of them are potentially radioactive but have not been observed to decay), which is probably related to the fact that 50 is a "magic number" of protons. 29 additional unstable isotopes are known, including the "doubly magic" tin-100 (Sn-100) (discovered in 1994) and tin-132 (Sn-132). The longest-lived radioisotope is Sn-126 with a half-life of 230,000 years. All other radioisotopes have half-lives less than a year. ' ' STRUCTURE OF Sn-102, Sn-104, Sn-106, Sn-108, Sn-110, Sn-112, Sn-114, Sn-116, Sn-118, Sn-120, Sn-122 Sn-124, AND Sn-126 WITH S = 0 For understanding the structure of this group (which includes the 7 stable isotopes with even number of nucleons) you must read my STRUCTURE OF Sn-112 . In the following diagram of Sn-100 with S =0 you see that the additional p49n49 and p50n5o make the symmetrical rectangles which contributes to the high symmetry of Sn-110. After a careful analysis we found that in the presence of an extra even number of neutrons with opposite spins one gets the structures of the above nuclides based on the Sn-100 with S = 0. ' ' ' DIAGRAM OF Sn-100 WITH S = 0' In this structure you see the six horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5, and -HP6 along with the horizontal squares of opposite spins like the -HSQ and +HSQ. Here the p45, n45, p47, n47, p46, n46, p48, and n48 make the symmetrical alpha particles of high symmetry. But you cannot see the additional p49n49 and p50n50 which makes the two symmetrical vertical rectangles with the n15p17 and the p16n18 respectively. Note that the p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' ' ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12........n12......p32' ' -HP6 p31.....n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' p47......n27.........p8..........n8.......p28....n48' ' -HP4 n45.....p27.......n7..........p7.......n28.........p46 ' ' n47......p25.........n6.........p6.......n26....p48' ' +HP3 p45.....n25……p5........n5……...p26.........n46 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2………p2........n22' ' +HP1 n21......p1........n1........p22 ' ' p37......n37 ' ' -HSQ n39......p39 ' Then in the presence of an even number of extra neutrons with opposite spins which fill the blank positions we get the unstable structures of the Sn-102....Sn-112 . Here the extra neutrons make two bonds per neutron but they do not lead to the stability because the small number of extra neutrons cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of Sn-112....Sn-124, the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. Whereas the two more extra neutrons of Sn-126 (in the absence of blank positions) make single bonds leading to the decay. ' ' STRUCTURE OF Sn-128, Sn-130, Sn-132, Sn-134, AND Sn-136, WITH S =0 Similarly the structure of the above unstable nuclides with even number of extra neitrons is based on the same structure of Sn-100 with S =0, but the more extra neutrons than those of Sn-126 make single bonds which lead to the decay.' ' ' ' STRUCTURE OF Sn-113, Sn-115, Sn-117, Sn-119 WITH S = +1/2 AND Sn-121 WITH S = +3/2 ' For understanding the structure of the above nuclides you must read my STRUCTURE OF Sn-115 . In this case using the diagram of Sn-100 with S =0 we see that the structure of them with S =+1/2 is given by adding extra neutrons with a total spin S =+1/2. For example the structure of the stable Sn-115 with S = +1/2 ,which has 15 extra neutrons, is giving by adding 8 extra neutrons of positive spins and 7 extra neutrons of negative spins giving a total S = +1/2. Moreover the structure of the unstable Sn-121 with S =+3/2 has two more extra neutrons of positive spins than those of the stable Sn-119 with S =+1/2. That is S = +1/2 + 2(+1/2) = +3/2 ' ' '''SRUCTURE OF Sn-101, Sn-103, Sn-105, Sn-107, Sn-109, AND Sn-111 ' After a careful analysis I found that the structure of the above unstable nuclides with odd number of extra neutrons is based on another structure of Sn-100 with S =+2. In this case the p37n37 of the diagram of Sn-100 with S =0 changes the spin from S = -1 to S = +1 giving S = +2 . Particularly it moves from -HSQ to +HSQ in order to make horizontal bonds with p38n38 Under this new structure of Sn-100 with S =+2, in the presence of the n51(+1/2 one gets the structure of the unstable Sn-101 with S =+5/2. That is S = +2 + 1(+1/2) = +5/2 Similarly the structures of Sn-103, Sn-105 and Sn-107 with S = +5/2 are based on the same new structure of Sn-100 with S= +2 because the extra neutrons of them give S = +1/2. For example the Sn-107 with S = +5/2 has 4 extra neutrons of positive spins and three extra neutrons of negative spins giving S = +1/2. That is S = +2 + 4(+1/2) +3(-1/2) = +5/2 On the other hand in the presence of two extra neutrons of positive spins than those of Sn-107 one gets the structure of Sn-109 with S = +7/2. That is S = +5/2 + 2(+1/2) = +7/2 Moreover in the structure of Sn-111 with S = +7/2 we have two more extra neutrons of opposite spins than those of Sn-109. ' ' '''STRUCTURE OF Sn-123, Sn-125, Sn-127, Sn-133, Sn-135, AND Sn-137 In the presence of such an odd number of extra neutrons I found that the structure of the above unstable nuclides is based on another structure of Sn-100 having S = -4. In this case the p38n38 and n40p40 of the +HSQ in the diagram of Sn-100 with S=0 change their spins from S = +2 to S = -2 giving S = -4 . Particularly they move from the +HSQ to -HSQ in order to make horizontal bonds with p37n37 and n39p39. Then in the presence of extra neutrons one gets the structure of the above nuclides. For example the Sn-123 with S = -11/2 (which has 23 extra neutrons), it has 13 extra neutrons of negative spins and 10 extra neutrons of positive spins. That is S = -4 + 13(-1/2) +10(+1/2) = -11/2 Whereas the Sn-137 with S = -5/2 (which has 37 extra neutrons), it has 20 extra neutrons of positive spins and 17 extra neutrons of negative spins. That is S = -4 + 20(+1/2) + 17(-1/2) = -5/2 . STRUCTURE OF Sn-99 WITH S = +9/2 ''' In the absence of one neutron I found that the structure of the above unstable nuclide is based on another structure of Sn-100 having not S = -4 but S = +4. In this case the p37n37 and n39p29 of the -HSQ in the diagram of Sn-100 with S=0 change their spins from S = -2 to S = +2 giving S = +4 . Particularly they move from the -HSQ to +HSQ in order to make horizontal bonds with p38n38 and n40p40. Then in the absence of one neutron with negative spin we get the structure of the above nuclide with S = +9/2. That is S = +4 - 1(-1/2) = +9/2 '''STRUCTURE OF Sn-129 AND Sn-131 with S = +3/2 In the presence of such an odd number of extra neutrons I found that the structure of the above unstable nuclides is based on another structure of Sn-100 having S = +2. In this case the p37n37 of the -HSQ in the diagram of Sn-100 with S=0 changes the spin from S = -1 to S = +1 giving S = +2 . Particularly it moves from the -HSQ to +HSQ in order to make horizontal bonds with n40p40. Then in the presence of extra neutrons we get the structure of the above nuclides. For example the Sn-131 with 31 extra neutrons has 30 extra neutrons of opposite spins giving S =0 and one extra neutron of negative spin. That is S = +2 + 0 + 1(-1/2) = +3/2 Category:Fundamental physics concepts